Mathematics - Application of Derivative Question with Solution | TestHub

$f\left(x\right)=\left(4a-3\right)\left(x+{\log }_{e}5\right)+2\left(a-7\right)\cot \left(\frac{x}{2}\right){\sin }^2\left(\frac{x}{2}\right),x\ne 2n\pi ,n\in N$

$f\left(x\right)=\left(4a-3\right)\left(x+{\log }_{e}5\right)+\left(a-7\right)\sin x$

$f^{\text{'}}\left(x\right)=\left(4a-3\right)\left(1\right)+\left(a-7\right)\cos x=0$

$\Rightarrow \cos x=\frac{3-4a}{a-7}$

$-1\le \frac{3-4a}{a-7}\le 1$

$\frac{3-4a}{a-7}+1\ge 0 \frac{3-4a}{a-7}\le 1$

$\frac{3-4a+a-7}{a-7}\ge 0 \frac{3-4a}{a-7}-1\le 0$

$\frac{-3a-4}{a-7}\ge 0 \frac{3-4a-a+7}{a-7}\le 0$

$\frac{3a+4}{a-7}\le 0 \frac{-5a+10}{a-7}\le 0$

$\frac{3a+4}{a-7}\le 0 \frac{5a-10}{a-7}\ge 0$

$\frac{3a+4}{a-7}\le 0 \frac{5\left(a-2\right)}{a-7}\ge 0$

$\alpha \in \left[-\frac{4}{3},7\right) \alpha \in (-\infty ,2]\cup (7,\infty )$

Hence, $\alpha \in \left[-\frac{4}{3},2\right]$